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Unformatted text preview: B ˙ ILKENT UNIVERSITY Department of Mathematics MATH 225, LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS, Solution of Homework set 1 # 5 U. Mu˘gan June 20, 2008 Homework problems from the 2 nd Edition, SECTION 3.1 3 (3) ) 2 Substruction of 3 / 2 times the first equation from the second eq. gives y/ 2 = 3 / 2 So, y = 3 and from the first equation (back substitution) x = 4. 4 (4) ) Substruction of 6 / 5 times the first equation from the second eq. gives 11 y/ 5 = 44 / 5 So, y = 4 and from the first equation x = 5. 6 (6) ) Substruction of 3 / 2 times the first equation from the second eq. gives 0 = 1. So, the system is inconsistent and hence no solution exists. 7 (7) ) Note that second eq. is 2 times the first equation. So we can choose y = t arbitrarily. Then the first eq. gives x = 10 + 4 t . 11 (11) ) First swap the first and second equations. Then substraction of twice the new first equation from the new second eq. gives y z = 7, and substraction 3 times the new first eq. from the third eq....
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This note was uploaded on 06/18/2010 for the course COMPUTER S Math 225 taught by Professor Yosumhoca during the Spring '10 term at Bilkent University.
 Spring '10
 YosumHoca

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